Lesson Ambiguous case SSA

Algebra ->  Algebra  -> Trigonometry-basics -> Lesson Ambiguous case SSA      Log On

Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!
Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

   

This Lesson (Ambiguous case SSA) was created by by Edwin McCravy(7325) About Me : View Source, Show
About Edwin McCravy: 
This is an illustration of the AMBIGUOUS case SSA, (side-side-angle) 
where you are asked to solve a triangle given two sides and a 
non-included angle.

Such a triangle may have 0 solutions, 1 solution, or 2 solutions. If 
it has 2 solutions we call them triangle ABC with sides a,b,c, and 
triangle A'B'C' with sides a',b',c'.

Suppose we are given that 

A=60 degrees, a=9, c=10

The first two figures below show how there could be two solutions, and 
the third figure below shows them put together with triangle A'B'C' inside of
triangle ABC, the blue arc shows that a and a' both equal 9 in length:

drawing%283600%2F11%2C400%2C-1%2C9%2C-1%2C10%2Ctriangle%280%2C0%2C7.7449489743%2C0%2C5%2C5sqrt%283%29%29%2C+%0D%0Alocate%28.5%2C.6%2C%2760%B0%27%29%2C+red%28arc%280%2C0%2C3%2C-3%2C0%2C60%29%29%2C+locate%281.1%2C4.5%2Cc=10%29%2C%0D%0Alocate%280%2C0%2CA%29%2C+locate%287.6%2C0%2CC%29%2C+locate%285%2C9.1%2CB%29%2C%0D%0Alocate%286.7%2C4.5%2Ca=9%29%29drawing%283600%2F11%2C400%2C-1%2C9%2C-1%2C10%2Ctriangle%280%2C0%2C2.550510257%2C0%2C5%2C5sqrt%283%29%29%2C%0D%0Alocate%28.5%2C.6%2C%2760%B0%27%29%2C+red%28arc%280%2C0%2C3%2C-3%2C0%2C60%29%29%2C+locate%281%2C4.5%2C%27c%27=10%27%29%2C%0D%0Alocate%280%2C0%2C%27A%27%27%29%2C+locate%282.550510257%2C0%2C%27C%27%27%29%2C+locate%285%2C9.1%2C%27B%27%29%2C%0D%0Alocate%284%2C4.5%2C%27a%27=9%27%29%2C+%0D%0Alocate%28.5%2C.6%2C%2760%B0%27%29%2C+red%28arc%280%2C0%2C3%2C-3%2C0%2C60%29%29%29drawing%283600%2F11%2C400%2C-1%2C9%2C-1%2C10%2Ctriangle%280%2C0%2C2.550510257%2C0%2C5%2C5sqrt%283%29%29%2C%0D%0Alocate%28.5%2C.6%2C%2760%B0%27%29%2C+red%28arc%280%2C0%2C3%2C-3%2C0%2C60%29%29%2C+locate%281%2C4.1%2C%27c%27=10%27%29%2C+locate%282.550510257%2C0%2C%27C%27%27%29%2C+locate%285%2C9.3%2C%27B_%26_B%27%27%29%2C%0D%0Alocate%284%2C4.5%2C%27a%27=9%27%29%2Ctriangle%280%2C0%2C7.7449489743%2C0%2C5%2C5sqrt%283%29%29%2C+%0D%0Alocate%28.5%2C.6%2C%2760%B0%27%29%2C+red%28arc%280%2C0%2C3%2C-3%2C0%2C60%29%29%2C+locate%281.1%2C4.7%2Cc=10%29%2C%0D%0Alocate%28-.3%2C.2%2C%27A_%26_A%27%27%29%2C+locate%287.6%2C0%2CC%29%2C+%0D%0A%0D%0Ablue%28arc%285%2C5sqrt%283%29%2C17.1%2C-18.3%2C250%2C290%29%29%2C%0D%0A%0D%0A%0D%0Alocate%286.7%2C4.5%2Ca=9%29%29


First solution               Second solution (maybe)

A = 60°                      A' = 60°
B =                          B' = 
C =                          C' = 
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10


We start with the law of sines:

sin%28A%29%2Fa = sin%28C%29%2Fc

sin%28%2760%B0%27%29%2F9 = sin%28C%29%2F10

9·sin(C) = 10·sin(60°)

sin(C) = %2810sin%28%2760%B0%27%29%29%2F9

sin(C) = .9622504486  <--- If this had been greater than 1, there
                           whould have been no solution.  If this had 
                           been exactly 1, there would have been 1 right
                           triangle angle solution.  But since it is less 
                           than 1, we can tell that there is either 1 or 2
                           solutions.  
                
If you use your calculator with the inverse sine,
you get 

C = 74.20683095°.

That is a correct value for angle C.  However there is another 
possible angle with that same sine, which is a second quadrant 
angle, and it is found by subtracting 74.20683095° from 180°.
We'll call it C':

C' = 180° - 74.20683095° = 105.793169°

So we put those two values in:

First solution               Second solution (maybe)

A = 60°                      A' = 60°
B =                          B' = 
C = 74.20683095°.            C' = 105.793169°
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10  

To find out whether there are 2 solutions or only 1,
we calculate the angles B and B', and see if both
are possibilities:

We calculate B by using 180°-A-C = 180°-60°-74.20683095° = 45.79316905°

We calculate B' by using 180°-A'-C' = 180°-60°-105.793169° = 14.20683095°

Since B' came out positive, we know that there are two solutions ABC and A'B'C',
[If B' had come out negative, there would have been only 1 solution ABC].

First solution               Second solution

A = 60°                      A' = 60°
B = 45.79316905°             B' = 14.20683095°
C = 74.20683095°.            C' = 105.793169°
a =  9                       a' =  9  
b =                          b' =  
c = 10                       c' = 10 

Now we just have to calculate sides b and b'

To calculate side b:

sin%28A%29%2Fa = sin%28B%29%2Fb

sin%28%2760%B0%27%29%2F9 = sin%28%2745.79316905%B0%27%29%2Fb 

b·sin(60°) = 9·sin(45.79316905°)  

b = %289%2Asin%28%2714.20683095%B0%27%29%29%2Fsin%28%2760%B0%27%29

b = 7.449489743

---

To calculate side b':

sin%28%27A%27%27%29%2F%27a%27%27 = sin%28%27B%27%27%29%2F%27b%27%27

sin%28%2760%B0%27%29%2F9 = sin%28%2745.79316905%B0%27%29%2F%27b%27%27 

b'·sin(60°) = 9·sin(45.79316905°)  

b' = %289%2Asin%28%2745.79316905%B0%27%29%29%2Fsin%28%2760%B0%27%29

b' = 2.550510257

So we end up with:

First solution               Second solution

A = 60°                      A' = 60°
B = 45.79316905°             B' = 14.20683095°
C = 74.20683095°             C' = 105.793169°
a =  9                       a' =  9  
b = 7.449489743              b' = 2.550510257  
c = 10                       c' = 10

Edwin
This lesson has been accessed 426 times.